Question

find the smallest number which is divided by 5,10,12,15,giving the remainder 2 but when divided by 7 leaves no remainder.​

Answer 1

Answer:

Given that the smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively.

28 - 8 = 20 and 32 - 12 = 20 are divisible by the required numbers.

Therefore the required number will be 20 less than the LCM of 28 and 32.

Prime factorization of 28 = 2 * 2 * 7

Prime factorization of 32 = 2 * 2 * 2 * 2 * 2

LCM(28,32) = 2 * 2 * 2 * 2 * 2 * 7

= 224.

Therefore the required smallest number = 224 - 20

= 204.

Verification:

204/28 = 28 * 7 = 196.

= 204 - 196

= 8

204/32 = 32 * 6 = 192

= 204 - 192

= 12.

Hope this helps!

Answer 2

5 because 5*1=5

5*2=10

5*3=15

5*2=10

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